Efficiently reverse the order of the words (not characters) in an array of characters

Question

Given an array of characters which forms a sentence of words, give an efficient algorithm to reverse the order of the words (not characters) in it.

Example input and

Answer 1

@Daren Thomas

Implementation of your algorithm (O(N) in time, O(1) in space) in D (Digital Mars):

#!/usr/bin/dmd -run
/**
 * to compile & run:
 * $ dmd -run reverse_words.d
 * to optimize:
 * $ dmd -O -inline -release reverse_words.d
 */
import std.algorithm: reverse;
import std.stdio: writeln;
import std.string: find;

void reverse_words(char[] str) {
  // reverse whole string
  reverse(str);

  // reverse each word
  for (auto i = 0; (i = find(str, " ")) != -1; str = str[i + 1..length])
    reverse(str[0..i]);

  // reverse last word
  reverse(str);
}

void main() {
  char[] str = cast(char[])("this is a string");
  writeln(str);
  reverse_words(str);
  writeln(str);
}

Output:

this is a string
string a is this

Answer 2

A solution in C/C++:

void swap(char* str, int i, int j){
    char t = str[i];
    str[i] = str[j];
    str[j] = t;
}

void reverse_string(char* str, int length){
    for(int i=0; i<length/2; i++){
        swap(str, i, length-i-1);
    }
}
void reverse_words(char* str){
    int l = strlen(str);
    //Reverse string
    reverse_string(str,strlen(str));
    int p=0;
    //Find word boundaries and reverse word by word
    for(int i=0; i<l; i++){
        if(str[i] == ' '){
            reverse_string(&str[p], i-p);
            p=i+1;
        }
    }
    //Finally reverse the last word.
    reverse_string(&str[p], l-p);
}

This should be O(n) in time and O(1) in space.

Edit: Cleaned it up a bit.

The first pass over the string is obviously O(n/2) = O(n). The second pass is O(n + combined length of all words / 2) = O(n + n/2) = O(n), which makes this an O(n) algorithm.

Answer 3

Efficient in terms of my time: took under 2 minutes to write in REBOL:

reverse_words: func [s [string!]] [form reverse parse s none]

Try it out: reverse_words "this is a string" "string a is this"

Answer 4

THIS PROGRAM IS TO REVERSE THE SENTENCE USING POINTERS IN "C language" By Vasantha kumar & Sundaramoorthy from KONGU ENGG COLLEGE, Erode.

NOTE: Sentence must end with dot(.) because NULL character is not assigned automatically at the end of the sentence*

 #include<stdio.h>
 #include<string.h>

int main()
{
char *p,*s="this is good.",*t;
int i,j,a,l,count=0;

l=strlen(s);

p=&s[l-1];

t=&s[-1];
while(*t)
   {
      if(*t==' ')
     count++;
     t++;
   }
   a=count;
  while(l!=0)
   {
for(i=0;*p!=' '&&t!=p;p--,i++);
   p++;

  for(;((*p)!='.')&&(*p!=' ');p++)
    printf("%c",*p);
  printf(" ");
  if(a==count)
   {
     p=p-i-1;
     l=l-i;
   }
  else
   {
     p=p-i-2;
     l=l-i-1;
   }

count--;
  }

return 0;  
}

Answer 5

Here is my answer. No library calls and no temp data structures.

#include <stdio.h>

void reverse(char* string, int length){
    int i;
    for (i = 0; i < length/2; i++){
        string[length - 1 - i] ^= string[i] ;
        string[i] ^= string[length - 1 - i];
        string[length - 1 - i] ^= string[i];
    }   
}

int main () {
char string[] = "This is a test string";
char *ptr;
int i = 0;
int word = 0;
ptr = (char *)&string;
printf("%s\n", string);
int length=0;
while (*ptr++){
    ++length;
}
reverse(string, length);
printf("%s\n", string);

for (i=0;i<length;i++){
    if(string[i] == ' '){
       reverse(&string[word], i-word);
       word = i+1;
       }
}   
reverse(&string[word], i-word); //for last word             
printf("\n%s\n", string);
return 0;
}

Answer 6

pushing a string onto a stack and then popping it off - is that still O(1)? essentially, that is the same as using split()...

Doesn't O(1) mean in-place? This task gets easy if we can just append strings and stuff, but that uses space...

EDIT: Thomas Watnedal is right. The following algorithm is O(n) in time and O(1) in space:

1. reverse string in-place (first iteration over string)
2. reverse each (reversed) word in-place (another two iterations over string)
   1. find first word boundary
   2. reverse inside this word boundary
   3. repeat for next word until finished

I guess we would need to prove that step 2 is really only O(2n)...